Differentiation (Part-38) | S N De

In the  previous article (Set Theory : Part-1), we  discussed  Short Answer Type Questions (1-6) of S. N. De Math Exercise -1 as a part of our S. N. De Math Solution Series. In this article , we will continue our journey to solve few more Short Ans Type Questions .

 7. Let $\,\, A=\{a,b,c,d,e\},\,\,B=\{a,c,e,g\},\,\,\text{and}\\ C=\{b,c,f,g\}. $ 

Verify that (i) $ (A \cup B)\cap C=(A \cap C)\cup(B \cap C)$

Sol. $ (A \cup B) = \{a,b,c,d,e\} \cup \{a,c,e,g\}\\ = \{a,b,c,d,e,g\} \\  (A \cup B)\cap C = \{a,b,c,d,e,g\} \cap \{b,c,f,g\}\\~~~~~~~~~~~~~~~~~~=\{b,c,g\} .......(1) \\ \text{Now,}\,\, A \cap C = \{a,b,c,d,e\} \cap \{b,c,f,g\}\\~~~~~~~~~~~~~~~~~~~~~= \{b,c\} \\ \text{and}\,\,  B \cap C = \{a,c,e,g\} \cap \{b,c,f,g\}= \{c,g\} \\ \therefore (A \cap C)\cup(B \cap C)= \{b,c\}\cup \{c,g\}\\=\{b,c,g\}.........(2)$

So, from (1) and (2), the result follows.

 (ii) $ (A \cap B)\cup C=(A \cup C)\cap(B \cup C)  $

Sol.  $ (A \cap B)= \{a,b,c,d,e\} \cap \{a,c,e,g\}=\{a,c,e\} \\  \text{and}  \\  (A \cap B) \cup C=  \{a,c,e\} \cup    \{b,c,f,g\}\\=\{a,b,c,e,f,g\}..........(1) \\    A \cup C = \{a,b,c,d,e\} \cup \{b,c,f,g\}\\= \{a,b,c,d,e,f,g\} \\ \text{and}\,\,  B \cup C = \{a,c,e,g\} \cup \{b,c,f,g\}\\= \{a,b,c,e,f,g\} \\ (A \cup C ) \cap (B \cup C )\\= \{a,b,c,d,e,f,g\} \cap \{a,b,c,e,f,g\}\\=\{a,b,c,e,f,g\}........(2)$

So, from (1) and (2), the result follows.

8(i) Let $S = \{1, 2, 3, 4, 5\}\,\,$ be the universal set and let $\,A=\{3, 4, 5\}\,$ and $\,\,B=\{1,4,5\}\,\,$ be two of its subsets. Verify :  $ (A \cup B)' = A'\cap B'\,\,$ (dash denotes complement).

Sol. Here, $ A \cup B= \{3, 4, 5\} \cup  \{1,4,5\}=\{1,3,4,5\} \\ \therefore ( A \cup B)'= S-( A \cup B)\\=\{1, 2, 3, 4, 5\}- \{1,3,4,5\}\\=\{2\} \cdots (1) \\ A'=S-A\\= \{1, 2, 3, 4, 5\}-\{3, 4, 5\}\\=\{1,2\} \\ B'=S-B\\= \{1, 2, 3, 4, 5\}-\{1, 4, 5\}\\=\{2,3\} \\ \therefore A' \cap B'= \{1,2\} \cap \{2,3\}\\=\{2\} \cdots (2)$ 

So, from (1) and (2) , the result follows.

8(ii) The set $\,\,S = \{1, 2, 3, , 12\}\,\,$ is to be partitioned into three sets $\,\,A, B, C\,\,$ of equal size. Thus $\,\, A \cup B \cup C=S, \\ A \cap B=B \cap C=C \cap A=\phi.$

Find the number of ways to partition $\,S.$

Solution: Here the given set $\,\,S = \{1, 2, 3, … , 12\}$

Now, in total there are 12 elements that have to be partitioned into three sets.

Therefore, each set gets $\,4\,$ elements.

Since, $A\cap B=B\cap C=A \cap C=ϕ,\,\,$ it means that all the three sets have different elements.

Hence, total number of ways to partition $\,S\,$ is : $\,\, {}^{12}C_4 \times {}^8C_4 \times {}^4C_4 \\=\frac{12!}{(12-4)! \times 4!} \times \frac{8!}{(8-4)! \times 4!} \times \frac{4!}{(4-4)! \times 4!}\\=\frac{12!}{(4!)^3} $

9.  If    $A=\{1,2,3,4\},\,B=\{2,3,4,5\},\\ C=\{1,3,4,5,6,7\}, \\ \text{find}\,\, (i) A-B ,\,(ii)  A-C \, $  and hence verify that $\,\, A- ( B \cap C)=(A-B)\cup (A-C).$ 

Sol. $(i) A-B=\{ x : x \in A \wedge x \not \in B\}\\=\{1,2,3,4\}-\{2,3,4,5\}\\=\{1\}\\ A-C= \{ x : x \in A \wedge x \not \in C\}\\=\{1,2,3,4\}-\{1,3,4,5,6,7\}\\=\{2\} \\ \therefore (A-B) \cup (A-C)\\=\{x : x \in (A-B) \vee x \not \in (A-C)\}\\=\{1\} \cup \{2\}\\=\{1,2\} \cdots (1)\\ \text{Also, we notice }\,\, (B \cap C)\\=\{ x : x \in B \wedge x  \in C\}\\=\{2,3,4,5\} \cap \{1,3,4,5,6,7\}\\=\{3,4,5\} \\ \therefore A-(B \cap C) = \{x: x \in A \wedge x \not \in (B \cap C)\} \\ =\{1,2,3,4\}-\{3,4,5\}\\=\{1,2\} \cdots (2)$

From (1) and (2), the result follows. 

10. If $S=\{1, 2, 4, 8, 16, 32\}\,\, $be the universal set and $\,A = \{1, 2, 8, 32\},\,\, B = \{4, 8, 32\}\,\,$ be two of its subsets, verify that,

$(i) (A^c)^c = A,\,(ii) (A \cap B)^c = A^c \cup B^c \\(iii) (A \cup B)^c = A^c \cap B^c$

Sol. We know , $A^c=\{x: x \not \in A\}\\=\{x: x \in S-A\}\\=\{1, 2, 4, 8, 16, 32\}- \{1, 2, 8, 32\} \\= \{4,16\} \\ \therefore (A^c)^c=\{x: x \not \in A^c \}\\=\{x: x \in S-A^c\}\\=\{1, 2, 4, 8, 16, 32\}-\{4,16\}\\=  \{1, 2, 8, 32\}=A \cdots (1)\\ \text{So, from (1), the result follows.}$

(ii)$ A \cap B= \{x: x \in A \wedge x \in B\}\\=\{8,32\} \\ \therefore (A \cap B)^c= \{x: x \not \in (A \cap B) \}\\=\{x: x \in S-(A \cap B)\}\\=\{1,2,4,16\} \cdots (1)\\ \text{Now,}\,\, A^c=\{x: x \not \in A\}\\=\{x: x \in (S-A)\}\\=\{4,16\} \\ \text{Also, similarly}\,\,B^c=\{x: x \not \in B\}\\=\{x: x \in (S-B)\}\\=\{1,2,16\} \\ \text{So,}\,\, A^c \cup B^c=\{1,2,4,16\} \cdots (2)$

So, from (1) and (2), the result follows.

$(iii) A \cup B= \{x: x \in A \vee x \in B\}\\=\{1,2,4,8,32\} \\(A \cup B)^c =\{x: x \not \in (A \cup B) \}\\ =\{x: x \in S-(A \cup B)\} =\{16\} \cdots (1) \\ A^c \cap B^c=\{x: x \in A^c \wedge x \in B^c\}\\= \{16\} \cdots (2) \\ \text{Hence from (1) and (2), the result follows.} $

11(i) If $P=\{a,b,c,d,e,f\} ,\,\, Q=\{a,c,e,f\} \,\, $ then prove that $ (P-Q) \cup (P \cap Q)=P $

Sol. $P-Q=\{x: x \in P \wedge x \not \in Q\}\\=\{a,b,c,d,e,f\}- \{a,c,e,f\}\\=\{b,d\} \\ P \cap Q=\{x: x \in P \wedge x  \in Q\}\\= \{a,b,c,d,e,f\} \cap \{a,c,e,f\}\\=\{a,c,e,f\} \\ (P-Q) \cup (P \cap Q)\\ =\{x: x \in (P-Q) \vee x \in (P \cap Q) \}\\=\{b,d\} \cup \{a,c,e,f\}\\=\{a,b,c,d,e,f\}=P$

11(ii) If $\,P=\{\theta: \sin{\theta}-\cos{\theta}= \sqrt{2} \cos{\theta} \},  \\ Q=\{\theta: \sin{\theta}+\cos{\theta}= \sqrt{2} \sin{\theta}\},$

 then show that,$\,\, P=Q. $

Sol. For that we have to prove 

$\,\, \sin{\theta}-\cos{\theta} =\sqrt{2} \cos{\theta}\cdots(1) \\ \Rightarrow  \sin{\theta}+\cos{\theta}= \sqrt{2} \sin{\theta} \\ \text{Now squaring both sides of (1),}\\ \sin^2\theta +\cos^2 \theta -2\sin \theta \cos \theta=2 \sin^2 \theta \\ \Rightarrow \sin^2 \theta =\cos^2\theta +2 \sin \theta \cos \theta \\ \Rightarrow \sin^2 \theta +\sin^2 \theta = \sin^2 \theta +\cos^2\theta +2 \sin \theta \cos \theta \\ \Rightarrow 2 \sin^2 \theta=( \sin \theta+ \cos \theta)^2 \\ \Rightarrow \sin \theta +\cos \theta =\sqrt{2} \sin \theta \cdots(2) \\ \text{From (1) and (2), we see that}\,\, x \in P \\   \Leftrightarrow x \in Q.$

Hence completes the proof.  

12.  Given $\,\,A=\{1,2,3,4,5\}$ and $\,\, B \cup C=\{3,4,6\}$; Find              

(i) $ (A \cap B) \cup (A \cap C) \\ (ii) (A-B) \cap (A-C) $

Sol. By distributive law, $\,\, (A \cap B) \cup (A \cap C) \\= A \cap (B \cup C)\\= \{1,2,3,4,5\} \cap  \{3,4,6\} \\= \{3,4\} \\  \text{For the second part,} \,\, (A-B) \cap (A-C) \\=(A \cap B^c) \cap (A \cap C^c)\\= A \cap (B^c \cap C^c)\\= A \cap (B \cup C)^c\\=A-(B\cup C)\,\, [\text{De Morgan Law}]\\= \{1,2,3,4,5\}-\{3,4,6\}\\=\{1,2,5\} $ 

13. Define $3$ sets  $\,P,Q,R\,$ such that $\,\, P \cap Q \neq \phi,  Q \cap R \neq \phi,  R \cap P \neq \phi \,\,\text{but}\,\, \\  P \cap Q \cap R  = \phi$

Sol. Suppose that $\,\, P=\{x,y\},Q=\{y,z\},R=\{x,z\} \\ \therefore P \cap Q =\{y\} \neq \phi,\\  Q \cap R =\{z\} \neq \phi,\\  R \cap P =\{x\}\neq \phi \\\,\,\text{but}\,\,  P \cap Q \cap R  \\= (P \cap Q) \cap R\\=\{y\} \cap \{z,x\} \neq \phi$

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